Upper figure: cylinder oil tank cross-section perpendicular to its horizontal axe. The vertical coordinate is the oil level in percentage.
Lower figure: graph of oil volume/max. volume (in %) versus oil level $l$ (in feet). The horizontal straight lines represent the area/volume ratio $A(l)/A(10)=V(l)/V(10)$ (in %) for every multiple of 10% from 0% to 100%.
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Since the tank radius is $5$, the oil level with respect to the bottom of
the tank is given by $l=5-5\cos \frac{\theta }{2}$, where $\theta $ is the
central angle as shown in the figure. The area of the tank cross section
filled with oil is
$$A(\theta )=\frac{25}{2}\theta -\frac{25}{2}\sin \theta $$
or
$$A(l)=25\arccos (\frac{5-l}{5})-\frac{25}{2}\sin (2\arccos (\frac{5-l}{5}))$$
The area ratio $A(l)/A(10)=V(l)/V(10)$ where $V(l)$ is the oil volume.
Let $f(l)$ denote this area ratio in percentage:
$$f(l)=\frac{100}{\pi }\arccos \left( 1-\frac{1}{5}l\right) -\frac{50}{\pi }\sin \left( 2\arccos \left( 1-\frac{1}{5}l\right) \right) $$
Here is the sequence of $f(l)$ values for $l=0,1,2,\ldots ,10$. The graph of $f(l)$ is shown above.
$f(0)=0$, $f(1)=5.2044$, $f(2)=14.238$, $f(3)=25.232$, $f(4)=37.353$, $f(5)=50$,
$f(6)=62.647$, $f(7)=74.768$, $f(8)=85.762$, $f(9)=94.796$, $f(10)=100$
Edit: One still needs to solve the nonlinear equation $f(l)-10k=0$ for $k=1,2,3,4,6,7,8,9$, e. g. by the Secant Method.
Edit 2: The problem of solving graphically, as shown in the secong figure, is that it would be very difficult, or impossible, to get the required accuracy of 0.01 (feet).
Update: The oil level marks (in feet) should be placed at
$0,1.57,2.54,3.40,4.21,$
$5,5.79,6.60,7.46,8.44,10$
corresponding to the oil volume percentage of
$0,10,20,30,40,$
$50,60,70,80,90,100$.
This calculation was based on the following $f$ function values:
$f(0)=0.0$, $f(1.5648)=10.0$, $f(2.5407)=20.0$, $f(3.40155)=30.0$, $f(4.21135)=40.0$
$f(5)=50.0$, $f(5.7887)=60.0$, $f(6.59845)=70.000$, $f(7.4593)=80.000$,
$f(8.4352)=90.000$, $f(10)=100.0$
Update 2 Figure of marks:
[Rearranged to show the sequence of editions and updates.]
Best Answer
Since you asked for a process, I'll give you two since I'm putting off doing my homework.
Convoluted Solution: So let's set up the problem. We know that the average value of any function can be given by $$\int_a^b \frac{y(b)-y(a)}{b-a} dt$$
We put our starting time at $0$, since that is when we begin looking at when the ball is thrown.
We know that we'll stop caring about the ball when it hits the ground, which means it's position is $0$, since we have the position function, we simply have to to set it to $0$ in order to find when it hits the ground. In other words, we have to solve $$85t -16t^2 = 0$$
From there, we have to find the velocity function, which is the derivative of the position function. In other words, $$\frac{dy}{dt} = \frac{d}{dt}[85t - 16t^2]$$
I'll let you solve these on your own, but here's the set up integral without the solution $$\int_0^b\frac{\frac{dy}{dt}(b)-\frac{dy}{dt}(0)}{b-0}dt$$
Easy Solution: Take a look at this picture of the position
From here it should be easy to figure out the average velocity of the ball.